Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

Gears & gear trains

Meshing gears
n1·N1 = n2·N2
Torque
T2 = T1·(N2/N1)·η
Power
P = T·ω, ω = 2π·n/60
Compound train
total ratio = product of the stage ratios

Torsion of shafts

Shear stress
τ = T·r/J
Solid shaft
J = πd⁴/32
Hollow shaft
J = π(D⁴ − d⁴)/32
Angle of twist
φ = TL/(GJ)
Power
P = T·ω = 2πnT/60

Method

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Mohr’s circle & stress transformation

Centre
C = (σx + σy)/2
Radius
R = √(((σx − σy)/2)² + τxy²)
Principal stresses
σ1,2 = C ± R
Principal direction
tan 2θp = 2τxy/(σx − σy)
Inclined plane
σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ

Method

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Axial stress & strain

Normal stress
σ = F/A
Strain
ε = ΔL/L
Hooke’s law
σ = E·ε
Elongation
ΔL = FL/(EA)
Thermal strain
εT = α·ΔT

Column buckling

Euler load
Pcr = π²EI/(KL)²
Effective length factor
K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress
σcr = Pcr/A
Slenderness
λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency
ωn = √(k/m), fn = ωn/(2π)
From static deflection
ωn = √(g/δst)
Critical damping
cc = 2√(km) = 2mωn
Damping ratio
ζ = c/cc
Damped frequency
ωd = ωn√(1 − ζ²)
Logarithmic decrement
δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
Springs in parallel
k = k1 + k2
Springs in series
1/k = 1/k1 + 1/k2

2Practice problems

1.Gears & gear trainsMedium
N1​ = 16, m = 4 mm

A 1.5 kW motor running at 2900 rpm drives a pinion with 16 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.

P = 1.5 kWn = 2900 rpmN1 = 16m = 4 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 300 mm720 rpmC = 500 mmθ = ?

An open belt connects a 120 mm pulley and a 300 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 300 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
T = 500 N·md = 25 mmL = 0.8 mφ = ?

A solid aluminium shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 500 N·md = 25 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 75 mm, di​ = 53 mm

A hollow shaft transmits 75 kW at 1000 rpm. Find the maximum shear stress τmax in the shaft.

P = 75 kWN = 1000 rpmdo = 75 mmdi = 53 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 80σy​ = 10τxy​ = −30MPa

A point in a loaded part has σₓ = 80 MPa, σᵧ = 10 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 80 MPaσy = 10 MPaτxy = −30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 20σy​ = 110τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 20 MPaσy = 110 MPaτxy = −20 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Ø 30 mmAluminium2 mP = 120 kN

Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).

P = 120 kNd = 30 mmL = 2 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
Ø 40 mmBrass0.5 mØ 12 mmBrass0.5 mP = 100 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 100 kNsegment 1: Ø40 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 4 md = 30 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 4 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
P = 70 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 70 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 70 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
tx40 mm3 cycles18.8 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 3 cycles later by a peak of 18.8 mm. Find the damping ratio ζ.

x0 = 40 mmx3 = 18.8 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 80m = 40 kg

A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 80 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 40 kgk1 = 80 kN/mk2 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 154.4 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s
T = P/ω = 1500/303.7 = 4.939 N·m

Pitch diameter

d = m·N = 4·16 = 64 mm

Tangential force

Ft = 2T/d = 2·4.939/0.064 = 154.4 N

Answer

Ft = 154.4 N
2.Gears & gear trains

Answer: θ = 159.3 °

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Belt angle

asin(0.18) = 10.37°

Angle of wrap

θ = 180 − 2·10.37 = 159.3°

Answer

θ = 159.3°
3.Torsion of shafts

Answer: φ = 22.99 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar second moment of area

J = π·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·800/(26000·38350) = 0.4012 rad
φ = 0.4012·180/π = 22.99°

Answer

φ = 22.99°
4.Torsion of shafts

Answer: τmax = 11.52 MPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s
T = P/ω = 75000/104.7 = 716.2 N·m

Polar second moment of area

J = π(75⁴ − 53⁴)/32 = 2.332 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 716197·37.5/2.332 × 10⁶ = 11.52 MPa

Answer

τmax = 11.52 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 46.1 MPa

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((80 − 10)/2)² + −30²) = 46.1 MPa

Answer

τmax = 46.1 MPa
6.Mohr’s circle & stress transformation

Answer: n = 4.026

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

στxyCσ1​σ2​

Principal stresses

C = 65 MPa, R = 49.24 MPa
σ1 = 114.2 MPa, σ2 = 15.76 MPa, σ3 = 0

Tresca

σmax − σmin = 114.2 MPa
n = 460/114.2 = 4.026

Answer

n = 4.026

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: ε = 2425 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·30²/4 = 706.9 mm²
σ = P/A = 120·1000/706.9 = 169.8 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 169.8/70000 = 0.002425 = 2425 µε

Answer

ε = 2425 µε
8.Axial stress & strain

Answer: σmax = 884.2 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A1 = π·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 79.58 MPa
σ2 = 884.2 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 884.2 MPa

Answer

σmax = 884.2 MPa
9.Column buckling

Answer: Pcr = 3.504 kN

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·39761/2800² = 3504 N = 3.504 kN

Answer

Pcr = 3.504 kN
10.Column buckling

Answer: d = 48.5 mm

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Required critical load

Pcr = n·P = 2.5·70000 = 175000 N
KL = 0.7·2500 = 1750 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·175000·1750²/(π³·200000))^(1/4) = 48.5 mm

Answer

d = 48.5 mm

Check: the slenderness KL/r is 144.3, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.04002

Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Logarithmic decrement

δ = (1/3)·ln(40/18.8) = 0.2517

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2517/√(39.48 + 0.06334) = 0.04002

Answer

ζ = 0.04002
12.Mechanical vibrations

Answer: fn = 10.07 Hz

Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 80 + 80 = 160 kN/m

Natural frequency

ωn = √(160000/40) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz

Answer

fn = 10.07 Hz

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